Proof that binomial $\binom{n}{k}$ is an integer for $0>n\in\mathbb{Z}$

Question

I have already looked at this proof here

Proving that $n \choose k$ is an integer

However I don't understand how I can use the Pascal identity for binommial coefficients if $n$ is a negative number.

I have had the idea to prove a Connection between the negative and the positive Counterpart.

I.e.

Maybe $\binom{n}{k}=-x\binom{-n}{k}$, where $x\in\mathbb{Z}$

I wrote down

$\frac{1}{k!}n(n-1)…(n-k+1)=x\cdot\frac{1}{k!}-n(-n-1)….(-n-k+1)$

What could be the $x$ ?

Thanks for reading.

Answer 1

For $n>0$ we have \begin{align} \binom{-n}k &=\frac{-n(-n-1)\cdots(-n-k+1)}{k!}\\ &=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}\\ &=(-1)^k\frac{(n+k-1)!}{k!(n-1)!}\\ &=(-1)^k\binom{n+k-1}{k} \end{align} hence it is an integer.

Answer 2

Alternative method:

Note that

$\binom{n}{k} = \binom{n-1}{k-1}+ \binom{n-1}{k} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N} \\ \Rightarrow \binom{n-1}{k} = \binom{n}{k}- \binom{n-1}{k-1} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N}$

Together with $\binom{n}{0}=1 \space \forall n \in \mathbb{Z}$ you can use this to prove by induction that $\binom{n}{k} \space 0>n\in\mathbb{Z}, k\in \mathbb{N}$ is the difference of two integers and so is always an integer.